Two cyclists 2

At the same time, two cyclists left the towns A and B at constant speeds. The first one going from town A to town B, and the second one from town B to town A. At one point of the trip they met. After they met, the first cyclist arrived at town B in 36min, the second cyclist arrived at town A in 25min. How long did it take for the first cyclist to reach the meeting point after leaving town A?

Correct result:

t1 =  30 min

Solution:

s1=v1 t1=v2 25 s2=v1 36=v2 t2 t1=t2  v1 t1=v2 25 v1 36=v2 t1  t1/36=25/t1 t12=25 36  t1=25 36=30 mins_{1}=v_{1} \ t_{1}=v_{2} \cdot \ 25 \ \\ s_{2}=v_{1} \cdot \ 36=v_{2} \ t_{2} \ \\ t_{1}=t_{2} \ \\ \ \\ v_{1} \ t_{1}=v_{2} \cdot \ 25 \ \\ v_{1} \cdot \ 36=v_{2} \ t_{1} \ \\ \ \\ t_{1}/36=25/t_{1} \ \\ t_{1}^2=25 \cdot \ 36 \ \\ \ \\ t_{1}=\sqrt{ 25 \cdot \ 36 }=30 \ \text{min}



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